Timeline for Connection 1-form on a principal bundle, uniqueness of the separation of tangent space?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 5, 2022 at 15:25 | comment | added | Alex | Why does the Ehresmann connection have to be invariant under the G-action? | |
| Sep 17, 2016 at 17:14 | comment | added | Deane Yang | Could you try an example? | |
| Sep 16, 2016 at 22:18 | comment | added | Chen Li | That makes great sense, thanks very much:) A nonrelavent but annoying question, how did you guys keep the math equations stay in the line ? I use $ at the head and end of my equitions, it works in the main post but it did not work in responds. | |
| Sep 16, 2016 at 21:51 | comment | added | Deane Yang | At each point $p \in P$, the connection $1$-form defines a linear map $T_pP \rightarrow \mathfrak{g}$. Condition 1 implies that the map is surjective, and therefore the dimension of the kernel is the dimension of $P$ minus the dimension of the Lie algebra, i.e., the dimension of the base. | |
| Sep 16, 2016 at 21:43 | comment | added | Chen Li | can you be more more specific about how ? Thanks very much | |
| Sep 16, 2016 at 21:40 | comment | added | Deane Yang | Condition 1 forces the kernel to have the right dimension. | |
| Sep 16, 2016 at 21:37 | comment | added | Chen Li | Yes, that is exactly what confuses me, so should we add a third condition, "the kernel of a connection one form is the horizontal space (bundle)?" | |
| Sep 16, 2016 at 20:44 | comment | added | Paul Siegel | If $\ker \omega_2$ is a subspace of $\ker \omega_1$ but the two kernels are not equal then the codimension of $\ker \omega_2$ must be strictly larger than the codimension of $\ker \omega_1$ which in turn is the dimension of $VP$. So in that case $\ker \omega_2$ cannot be complementary to $VP$ in $TP$ and hence $\omega_2$ cannot be a connection 1-form. | |
| Sep 16, 2016 at 20:30 | comment | added | Chen Li | (Please ignore the previous comment) Thanks very much for you detailed explanation, that is really helpful, but I still have one question. Let's stick to the first definition, assmue there are two connection 1-forms $${\omega _1}$$ and $${\omega _2}$$ which both satisfy $ \omega ({A^\# }) = A$ and $$\ker {\omega _1} = {H_u}P$$. if the kernel of the $${\omega _2}$$ is only a subspace of $$\ker {\omega _1} $$ or $$ \ker {\omega _2} \subset {H_u}P$$, $${\omega _2}$$ still satisfies the second condition,right ? so $${\omega _2}$$ is still a connection one form ? | |
| Sep 16, 2016 at 20:20 | comment | added | Chen Li | Thanks very much for you detailed explanation, that is really helpful, but I still have one question. Let me put it this way: | |
| Sep 16, 2016 at 18:03 | history | answered | Paul Siegel | CC BY-SA 3.0 |